What is the tangent of f(x)=(1+x)^{4/x},\at x=1 ?

The tangent of f(x)=(1+x)^{4/x},\at x=1 is y=64(-ln(2)+1/2)x+64ln(2)-16

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tangent of f(x)=(1+x)^{4/x},\at x=1

tangent of

f (x) = (1 + x)^{\frac{4}{x}}, at x = 1

y = 64 (- ln (2) + \frac{1}{2}) x + 64 ln (2) - 16

Solution steps

Find the tangent point:

(1, 16)

Find the slope of

f (x) = (1 + x)^{\frac{4}{x}} : \frac{df ( x )}{d x} = 4 (- \frac{ln ( 1 + x )}{x ^{2}} + \frac{1}{x ( 1 + x )}) (1 + x)^{\frac{4}{x}}

Compute the slope of

f (x) = (1 + x)^{\frac{4}{x}}

x = 1 : m = 64 (- ln (2) + \frac{1}{2})

Find the line with slope m=

64 (- ln (2) + \frac{1}{2})

and passing through

(1, 16) : y = 64 (- ln (2) + \frac{1}{2}) x + 64 ln (2) - 16

y = 64 (- ln (2) + \frac{1}{2}) x + 64 ln (2) - 16

(6, - \frac{7 π}{6})

f (x) = e^{x} ln (x)

4 - y = 2 x

(- 1, 0)

f (x) = 3

What is the tangent of f(x)=(1+x)^{4/x},\at x=1 ?
The tangent of f(x)=(1+x)^{4/x},\at x=1 is y=64(-ln(2)+1/2)x+64ln(2)-16